Question

Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis. y = 32 − x2, y = x2; about x = 4

Answer 1

Answer:

Required volume is $\frac{4864\pi}{3}$ unit.

Step-by-step explanation:

Given equations of curves,

$y=32-x^2$, $y=x^2$

substitute second in first we will get,

$x^2=36-x^2\implies x^2=16\implies x=\pm 4$

When $x=4, y=16$ and $x=-4, y=16$. Thus both curves intersect at the points (4,16),(-4,16). And thus $-4$.

Considering width of the representative rectangle as $\Delta x$ which is parallel to x-axis because of considering cylindrical shell. Therefore volume of the shell is given by,

$V_{Shell}=(\textit{Length of box})\times (\textit{Width of box})\times (\textit{Thikness of box})$

Now,

Length of generating box=Length of $\Delta x$ from line x=4(axis of revolution)=circumference of the shell=(4-x)

Width of box=$(36-x^2)-x^2$

Hence,

$V_{Shell}$

$=\int_{-4}^{4}2\pi (4-x)(36-2x^2)dx$

$=2\pi\int_{-4}^{4}(144-36x-8x^2+2x^3)dx$

$=2\pi\{144\big[x\big]_{-4}^{4}-18\big[x^2\big]_{-4}^{4}-\frac{8}{3}\big[x^3\big]_{-4}^{4}+\frac{1}{2}\big[x^4\big]_{-4}^{4}\}$

$=\frac{4864\pi}{3}$

which is required volume of generating area.